$G$ of prime order can’t embed into symmetric groups of degree smaller than $|G|$. Proof verification.

Question

As a test for my understanding of actions, I thought the following. Could you verify it down to the resulting claim, please?

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Let $G$ be a finite group. A homomorphism $\varphi: G \rightarrow S_m$ (symmetric group of some degree $m$) is equivalent to an action of $G$ on a set $X$ of $m$ elements. Its kernel is:

$$\operatorname{ker}\varphi=\lbrace g \in G \mid \operatorname{Fix}(g)=X\rbrace$$

where $\operatorname{Fix}(g):=\{x \in X \mid g\cdot x=x\}$. Therefore:

\begin{alignat}{1}
\operatorname{ker}\varphi=\lbrace e \rbrace &\Leftrightarrow |\operatorname{Fix}(g)|<|X|, \forall g \in G \setminus\lbrace e\rbrace \\
&\Rightarrow |\mathcal{O}||G|=\sum_{g \in G}|\operatorname{Fix}(g)|<|G||X| \\
&\Rightarrow |\mathcal{O}|<|X|\\
\tag 1
\end{alignat}

where $\mathcal{O}$ is the set of action's orbits. Therefore, $\varphi$ injective $\Rightarrow \exists x \in X$ s.t. $|O(x)|>1 \Rightarrow$ $\exists x \in X$ s.t. $|\operatorname{Stab}(x)|<|G|$ (Orbit-Stabilizer Theorem). If, in addition, $G$ has prime order, then $\exists x \in X$ s.t. $|\operatorname{Stab}(x)|=1$, and finally $\exists x \in X$ s.t. $|O(x)|=|G|$; but then $|X| \ge |G|$.

So, provided that so far so good:

Claim: If $G$ has prime order $p$ and $G \hookrightarrow S_m$, then $m \ge p$.

In words, $G$ of prime order can't embed into symmetric groups of degree smaller than $|G|$. (Said differently, Cayley's one is the "sharpest" embedding we can have in this case.)

Answer 1

Note that if $|G| = p > m$ is prime, then $p \not\mid |S_m| = m!$. Therefore there cannot be an injective homomorphism $G \to S_m$, because its image would be a subgroup of order $p$, contradicting Lagrange's theorem.

Edit: I realized only after answering that this is a proof verification question: so to add, yes, your proof is correct, although it is much more complex than necessary (as demonstrated by the above, much shorter proof).